percentage yeild

Start with the end of the problem and work your way backwards.We know the volume and the concentration of the HCl used in the final titration. If we work in dm3, mol/dm3 and moles then every volume we use, we will divide by 1000 to ensure that the units are consistent.So we can work out the number of moles of HCl used in the final titration by saying that concentration = moles/volume so (20/1000) x 0.1 = 0.002mol. of HCl.By looking at the final titration equation we know that one mole of HCl reacts with the 25cm3 of unreacted NaOH solution. So therefore we know that we must have used 0.002 moles of NaOH in this reaction. We also know that we used 25cm3 or (25/1000)dm3 of NaOH so we can now work out the concentration of the NaOH solution made up from the unreacted NaOH. Concentration = 0.002mol./0.025dm3 = 0.08mol/dm3So we now know that the concentration of the solution made up from the unreacted NaOH in the reaction was 0.08mol/dm3.We can now work out the number of moles of unreacted NaOH we had. We know that the initial solution was made up to 250cm3 so if we say that moles = volume x concentration then moles of NaOH is 0.25dm3 x 0.08mol/dm3 which gives us 0.02moles of unreacted NaOH.We now must look at the start of the question which gives us the initial volume and concentration of NaOH used. We can use this to determine the number of moles of NaOH that we used in the reaction.Moles = volume x concentration = 0.025dm3 x 2mol/dm3 = 0.05 moles.So we know that 0.05 moles of NaOH were used in the reaction and that 0.02 moles of NaOH did not react.Thus we can say that 0.03 (0.05-0.02) moles of NaOH fully reacted with the ammonium sulphate.Looking at the equation there is a 1:2 mole ratio. So 1mole of ammonium sulphate reacts with 2 moles of NaOH. This means that there are half the number of moles of ammonium sulphate that fully react.So 0.03/2 = 0.015mol. of ammonium sulphate.So we can now work out the mass of ammonium sulphate that reacted by using the equation moles = mass/mr or mass = moles x mr.Mr (relative molecular mass) of ammonium sulphate = 132.079g/mol.So mass of ammonium sulphate used = 132.079g/mol x 0.015mol. = 1.981g.We had 4.00g of lawn sand/ammonium sulphate in the mixture and we now know that 1.981g of ammonium sulphate were in that mixture.So the percentage of ammonium sulphate in the mixture must be 1.981/4.00 x 100 = 49.53% Hope that helps!!